Solve for n log base 2 of n = log base 2 of 2n+6

$\log_{2} (n) = \log_{2} (2 n) + 6$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{2} (n) - \log_{2} (2 n) = 6$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{n}{2 n}) = 6$

Step 3

Reduce the expression $\frac{n}{2 n}$ by cancelling the common factors.

Cancel the common factor.

$\log_{2} (\frac{n}{2 n}) = 6$

Rewrite the expression.

$\log_{2} (\frac{1}{2}) = 6$

Step 4

Rewrite $\frac{1}{2}$ as $1 \cdot 2^{- 1}$ .

$\log_{2} (1 \cdot 2^{- 1}) = 6$

Step 5

Rewrite $\log_{2} (1 \cdot 2^{- 1})$ as $\log_{2} (1) + \log_{2} (2^{- 1})$ .

$\log_{2} (1) + \log_{2} (2^{- 1}) = 6$

Step 6

Use logarithm rules to move $- 1$ out of the exponent.

$\log_{2} (1) - \log_{2} (2) = 6$

Step 7

Logarithm base $2$ of $2$ is $1$ .

$\log_{2} (1) - 1 \cdot 1 = 6$

Step 8

Multiply $- 1$ by $1$ .

$\log_{2} (1) - 1 = 6$

Step 9

Logarithm base $2$ of $1$ is $0$ .

$0 - 1 = 6$

Step 10

Subtract $1$ from $0$ .

$- 1 = 6$

Step 11

The equation is never true.

No solution